A rock was thrown from the top of a building that is 192 feet tall. it reached its maximum height of 256 feet 2 seconds after it was thrown. after how many seconds will the rock hit the ground?

Answer 1

Hi Jadarius,

The height h of a projectile above the ground after t seconds is given by :

h(t) = - (1 / 2) g t ² + v_o t + h_o

g is the acceleration due to gravity which on earth is approximately equal to 32 feet / s²

then h(t) = - (1 / 2)(32) t ² + v_o t + h_o

h(t) = - 16 t ² + v_o t + h_o

Since the rock was thrown from an initial height of 192 feet then h_o=192

the rock reached its maximum height of 256 feet after 2 seconds

then h(t) = - 16 t ² + v_o t + 192

For a quadratic function of the form h = a t ² + b t + c, the vertex is located at t = - b / 2a

and since a is negative, the maximum value of h occur at t = - b / 2a (vertex of the graph of h(t)).

since the rock reached its maximum height of 256 feet after 2 seconds

then 2 = - b / 2(-16)

2 = - b / -32

so b = 64

which means our initial velocity v_o = 64

then our equation becomes h(t) = - 16 t ² + 64 t + 192

to check if the equation is correct we can simply plug in t = 2 and see if it will give us the maximum height of 256:

h(t) = - 16 (2) ² + 64 (2) + 192 = = 256

so now we can find after how many seconds(t=?) the rock hit the ground (height = 0)

- 16 t ² + 64 t + 192 = 0

so to solve we need to factor out -16 as a GCF and divide the whole equation by -16 which will result in:

t ² - 4 t - 12 = 0

now we can factor the trinomial on the left side of the equation:

( t - 6 ) ( t + 2 ) = 0

t - 6 = 0 or t + 2 = 0

+6 +6 - 2 -2

t = 6 t = -2 ( not possible since time cannot be negative)

Now to check our answer we can simply plug in t = 6 in the equation - 16 t ² + 64 t + 192 = 0

-16 (6)² + 64(6) + 192 = 0

0 = 0

so the rock will hit the ground after 6 Seconds

I hope this helps, please feel free to ask any questions regarding this solution!

Best Regards.

Answer 2

The equation of motion of the ball is given by:

-16t^2 + v0t + x0

The first term is the acceleration due to gravity

v0 is the initial velocity

x0 is the starting height, which is 192 feet.

By taking the first derivative you get the velocity of the ball as a function of time:

v=v0-32t

The ball will reach the maximum when this velocity is zero which happens at t=2s.

So 0=v0-2*32, or v0=64

So the equation of motion is:

position = -16t^2 + 64t + 192 = -16(t^2-4t-12)

We want to know when it will hit the ground, i.e. position = 0

So t^2 - 4t - 12 = 0

This factors to (t-6)(t+2)=0

which has a solution t=6s or t=-2s. Since you cannot have negative time, the answer is it hits the ground at time = 6s.